# Kernel of a homomorphism from a free group into $\mathbb{Z}$.

Let $F$ be a non-abelian free group, that is, a free group of rank at least $2$, and let $\phi: F \rightarrow \mathbb{Z}$ be a nontrivial group homomorphism. How to prove that the kernel of $\phi$ is not finitely generated?

There is a useful result you could use to prove this.

Let $F$ be a free group and $E \leq F$ with $|F:E| = \infty$. Suppose that $\exists \: \{1\} \neq N \unlhd G$ with $N \leq E$. Then the rank of $E$ is infinite.

The proof is as follows and assumes familiarity with the theory of Schreier generators of subgroups of free groups.

Let $F$ be free on $X$, let $U$ be a Schreier transversal of $E$ in $F$ and, for $g \in F$, denote the element in $U \cap Eg$ by $\overline{g}$.

Let $1 \neq w = a_1 \cdots a_l \in N \leq E$, with $a_i \in X^{\pm 1}$. For $u \in U$, $Euw = Euwu^{-1}u = Eu$, since $uwu^{-1} \in N \leq E$. So $\overline{uw} = u$, and $uw \not\in U$, so there is a least $k$ such that $ua_1 \cdots a_k \notin U$. Let $u_k := ua_1 \cdots a_{k-1}$. Then $u_k \in U$ and $u_ka_k \notin U$, so $u_ka_k\overline{u_ka_k}^{-1}$ is not trivial. Since $U$ is infinite and $l$ is fixed, there is an infinite subset $V$ of $U$ and a fixed $k$ with $1 \le k \le l$, such that $k$ is minimal with $u_ka_k \notin U$ for all $u \in V$. Then $\left\{ u_ka_k\overline{u_ka_k}^{-1} : u \in V \right\}$ is an infinite subset of the set of Schreier generators of $E$, and hence $E$ has infinite rank.

For a completely different proof, that uses only the universal property of free groups, you could factor your map $\phi:F \to {\mathbb Z}$ as $\phi:F \to H \to {\mathbb Z}$, where $H$ is the wreath product ${\mathbb Z} \wr {\mathbb Z}$, and observe that the kernel of $H \to {\mathbb Z}$ is infinitely generated.

• I know this was a question from long ago, but how does the map $F \to \mathbb{Z}$ factor into $F \to H \to \mathbb{Z}?$ Apr 6, 2018 at 14:30
• Write $H = \langle a,b \rangle$ with $b$ in the base group of the wreath product, and define the map $H \to {\mathbb Z}$ by $a \mapsto 1$, $b \mapsto 0$. Now find a free basis $x_1,x_2,\ldots$ of $F$ such that $\phi(x_1) = 1$ and $\phi(x_i)=0$ for $i>1$, and map $F \to H$ by $x_1 \mapsto a$, $x_i \mapsto b$ for $i>1$. Apr 6, 2018 at 15:06
• You seem to be implying that any such $\phi$ is surjective since it hits $1$. Is this a general fact? May 20, 2022 at 11:27
• Ah, I see what you mean. But for any nontrivial map $\phi: F \to {\mathbb Z}$, the image is isomorphic to ${\mathbb Z}$, so I can assume that $\phi$ is surjective. May 20, 2022 at 12:40

This follows since the rank of any infinite-index subgroup of a nonabelian free group is infinite. One way to prove this is by covering space theory. An infinite index subgroup corresponds to a cover of a wedge of circles where every point has infinitely many preimages. Then, if the cover is not a tree, there is at least one loop, $L$. Acting on $L$ by the infinite group of deck transformations gives infinitely many copies of $L$, but some of them may intersect or even coincide. However, since each loop has finitely many edges, only finitely many copies of $L$ can intersect $L$. So we can produce infinitely many disjoint loops. Thus the graph has infinite rank.

• I think there might be a simpler proof, using elementary concepts. Anyway, thanks Jim Conant for the interesting answer. I am thinking about it.
– rla
Apr 19, 2012 at 13:05